Chapter 3.4, Problem 3.10E

a.

To determine

Find the covariance $s_{xy}$ using bivariate data.

Answer to Problem 3.10E

  $s_{xy}=-2$

Explanation of Solution

Given:

The given bivariate data is $\left(1,6\right),\left(3,2\right),\text{and}\left(2,4\right)$ .

Calculation:

First determine ${\displaystyle \sum x_i}{y}_i$ , ${\displaystyle \sum x_i}$ and ${\displaystyle \sum y_i}$ .

  $x_i\to$ first coordinates of the ordered pairs.

  $y_i\to$ second coordinates of the ordered pairs.

  $\begin{array}{l}{\displaystyle \sum x_i}{y}_i=1\cdot 6+3\cdot 2+2\cdot 4=20\\ {\displaystyle \sum x_i}=1+3+2=6\\ {\displaystyle \sum y_i=6+2+4=12}\end{array}$

For covariance $s_{xy}$ using formula $s_{xy}=\frac{{\displaystyle \sum x_i{y}_i}-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{n}}{n-1}$ .

Where $n=$ number of ordered pairs.

  $n=3$

  $\begin{array}{l}{s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}\\ =\frac{20-\frac{72}{3}}{3-1}\\ =\frac{20-24}{2}\\ =\frac{-4}{2}\\ =-2\end{array}$

Hence $s_{xy}=-2$ .

b.

To determine

Find the correlation coefficient $r$ using bivariate data.

Answer to Problem 3.10E

The value of correlation coefficient $r=-1$ .

Explanation of Solution

Given:

The given bivariate data is $\left(1,6\right),\left(3,2\right),\text{and}\left(2,4\right)$ .

Calculation:

First determine ${\displaystyle \sum x_i}{y}_i$ , ${\displaystyle \sum x_i}$ and ${\displaystyle \sum y_i}$ .

  $x_i\to$ first coordinates of the ordered pairs.

  $y_i\to$ second coordinates of the ordered pairs.

  $\begin{array}{l}{\displaystyle \sum x_i}{y}_i=1\cdot 6+3\cdot 2+2\cdot 4=20\\ {\displaystyle \sum x_i}=1+3+2=6\\ {{\displaystyle \sum x_i}}^2=1^2+3^2+2^2=14\\ {\displaystyle \sum y_i=6+2+4=12}\\ {\displaystyle \sum y_i{}^2=6^2+2^2+4^2=56}\end{array}$

Find the sample variance $s^2$ using the formula $s^2=\frac{{\displaystyle \sum x_i{}^2}-\frac{{\displaystyle \sum x_i{}^2}}{n}}{n-1}$

  $\begin{array}{l}{s}_x{}^2=\frac{14-\frac{6^2}{3}}{3-1}=\frac{14-12}{2}=1\\ s_y{}^2=\frac{56-\frac{{12}^2}{3}}{3-1}=\frac{56-48}{2}=4\end{array}$

Find sample standard deviation.

  $\begin{array}{l}{s}_x=\sqrt{1}=1\\ s_y=\sqrt{4}=2\end{array}$

For covariance $s_{xy}$ using formula $s_{xy}=\frac{{\displaystyle \sum x_i{y}_i}-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{n}}{n-1}$ .

Where $n=$ number of ordered pairs.

  $n=3$

  $\begin{array}{l}{s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}\\ =\frac{20-\frac{72}{3}}{3-1}\\ =\frac{20-24}{2}\\ =\frac{-4}{2}\\ =-2\end{array}$

Find the correlation coefficient $r$ using the formula $r=\frac{s_{xy}}{s_x{s}_y}$ .

  $r=\frac{-2}{1\cdot 2}=-1$

Hence the value of correlation coefficient $r=-1$ .

c.

To determine

Find the regression line using the given points.

Answer to Problem 3.10E

The regression line is $y=8-2x$ .

Explanation of Solution

Given:

The given bivariate data is $\left(1,6\right),\left(3,2\right),\text{and}\left(2,4\right)$ .

Calculation:

First determine ${\displaystyle \sum x_i}{y}_i$ , ${\displaystyle \sum x_i}$ and ${\displaystyle \sum y_i}$ .

  $x_i\to$ first coordinates of the ordered pairs.

  $y_i\to$ second coordinates of the ordered pairs.

  $\begin{array}{l}{\displaystyle \sum x_i}{y}_i=1\cdot 6+3\cdot 2+2\cdot 4=20\\ {\displaystyle \sum x_i}=1+3+2=6\\ {{\displaystyle \sum x_i}}^2=1^2+3^2+2^2=14\\ {\displaystyle \sum y_i=6+2+4=12}\\ {\displaystyle \sum y_i{}^2=6^2+2^2+4^2=56}\end{array}$

Find the sample variance $s^2$ using the formula $s^2=\frac{{\displaystyle \sum x_i{}^2}-\frac{{\displaystyle \sum x_i{}^2}}{n}}{n-1}$

  $\begin{array}{l}{s}_x{}^2=\frac{14-\frac{6^2}{3}}{3-1}=\frac{14-12}{2}=1\\ s_y{}^2=\frac{56-\frac{{12}^2}{3}}{3-1}=\frac{56-48}{2}=4\end{array}$

Find sample standard deviation.

  $\begin{array}{l}{s}_x=\sqrt{1}=1\\ s_y=\sqrt{4}=2\end{array}$

For covariance $s_{xy}$ using formula $s_{xy}=\frac{{\displaystyle \sum x_i{y}_i}-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{n}}{n-1}$ .

Where $n=$ number of ordered pairs.

  $n=3$

  $\begin{array}{l}{s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}\\ =\frac{20-\frac{72}{3}}{3-1}\\ =\frac{20-24}{2}\\ =\frac{-4}{2}\\ =-2\end{array}$

Find the correlation coefficient $r$ using the formula $r=\frac{s_{xy}}{s_x{s}_y}$ .

  $r=\frac{-2}{1\cdot 2}=-1$

Find the slope $b$ using the formula $b=r\cdot \frac{s_y}{s_x}$

  $b=-1\cdot \frac{2}{1}=-2$

Find the value of y-intercept.

  $\begin{array}{l}a=\overline{y}-b\overline{x}=\frac{{\displaystyle \sum y_i}}{n}-b\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{12}{3}-\left(-2\right)\frac{6}{3}=4+2\cdot 2=8\end{array}$

Regression line is $y=a+bx$

  $y=8-2x$

Hence the regression line is $y=8-2x$ .

d.

To determine

Draw a scatterplot for the given points.

Answer to Problem 3.10E

The scatterplot is

  

Explanation of Solution

Given:

The given bivariate data is $\left(1,6\right),\left(3,2\right),\text{and}\left(2,4\right)$ .

Calculation:

The equation is $y=8-2x$ .

The scatterplot for the given points are.